Infinitely Small Sequence

In other words, if for every `epsilon>0` there is number `N=N_(epsilon)`, such that for `n>N`, `|x_n-0|=|x_n|<epsilon.`

We can reformulate definition as follows: sequence `x_n` is infinitesimal if its absolute value becomes less than some specified `epsilon>0`, starting with some number.

If we now return to the variant `x_n` that has limit `a`, then difference `alpha_n=x_n-a` will be infinitesimal, because by definition of limit`|alpha_n|=|x_n-a|<epsilon`. And vice versa, if `alpha_n` is infinitesimal then `x_n->a`.

This leads to the following fact.

Fact. `x_n` has limit `a` if and only if `alpha_n=x_n-a` is infinitesimal.

Therefore, if `x_n->a` then `x_n` can be represented as `x_n=a+alpha_n`.

All variants are infinitesimal because `|x_n-0|=|x_n|=|1/n|<epsilon` when `n>1/epsilon`. Therefore, we can take (recall that `N` is natural number) `N=N_epsilon>[1/epsilon]`, where `[x]` is a floor function.

You see that they are infinitesimal (so their limit is 0), but they behave differently: first is always greater 0, second is always less than 0, third alternates sign.

Example 2. Consider sequence `x_n=(2+(-1)^n)/n`.

Corresponding list is `{1,3/2,1/3,3/4,1/5,1/2,...}`.

By triangle inequality we have that `|x_n|=|2/n+((-1)^n)/n|<=|2/n|+|((-1)^n)/n|=|2/n|+|1/n|=|3/n|`.

Using Bernoulli inequality `gamma^n>1+n(gamma-1)` where `n` is natural and `n>1`, `gamma>1`, with `gamma=a^(1/n)`, we obtain that `(a^(1/n))^n>1+n(a^(1/n)-1)` or `a^(1/n)-1<(a-1)/n`.

This means that `|x_n-1|=root(n)(a)-1<(a-1)/n<epsilon` when `n>N_epsilon=[(a-1)/epsilon]`.

However, we can go another way.

Inequality `|x_n-1|=a^(1/n)-1<epsilon` is equivalent to the inequality `1/n<log_a(1+epsilon)` or `n>1/(log_a(1+epsilon))`, so it holds when `n>N_epsilon=[1/(log_a(1+epsilon))]`.

As can be seen two ways of thinking led us to the two different expressions for `N_epsilon`. For example, when `a=10,epsilon=0.01` we obtain that `N_(0.01)=9/(0.01)=900` according to the first way and `N_(0.01)=[1/(0.00432....)]=231` according to the second way. Second way gave us the smallest of all possible values for `N_(0.01)` because `10^(1/231)=1.010017` is different from 1 on more than `epsilon=0.01`.

Same will be in general case, because when `a<=1/(log_a(1+epsilon))` we have that `a^(1/n)-1>=epsilon`.

However, we are not interested in finding the smallest value of `N_epsilon` if we only want to find limit. We are interested in finding such `N_epsilon` that inequality will hold for all `n>=N_epsilon`. We don't care is it smallest value or not.

Example 6. Important example of infinitesimal is sequence `alpha_n=q^n` where `|q|<1`.

To prove that `alpha_n->0` consider inequlaity `|alpha_n|=|q|^n<epsilon`. This inequality is equivalent to the inequality `lg(|q|^n)<lg(epsilon)` or `nlg|q|<lg(epsilon)` i.e. `n>(lg(epsilon))/(lg|q|)`.

Therefore, if we take (considering `epsilon<1`) `N_epsilon=[(lg(epsilon))/(lg|q|)]` then for `n>N_epsilon` above inequality will hold.

Similarly it can be shown that variant `beta_n=A*q^n`, where `|q|<1` and A is constant, is also infinitesimal.